There are TWO 2's right? Yes count them.Originally Posted by OnceDear
Chance of rolling a 2 on the first die is 1/12.
Now there are 11 [eleven] faces remaining. How many 2's? 1? Chance of a 2? 1/11.
Chance of rolling a 2-2 with 12 faces is 1/(11*12) = 1/132
Originally Posted by RS__
I'm Beginning to See the Light...
Would Archbishop Desmond Tutu (TWO-TWO) be the consultant of last resort on this matter?
AgreedOriginally Posted by RS__
No. Once you rolled the first die, 5 of those faces became inelligableOriginally Posted by RS__
Chance for second die is 1/6
1/12 X 1/6 = 1/72
Which is exactly half of 1/36. Coincidence? I think not.
Last edited by OnceDear; 05-13-2015 at 04:21 PM.
Helleluia.Originally Posted by kewlI'm Beginning to See the Light...
If you say (and I quote Pharoah here), "If you roll [present tense] the dice, and at least one of the dice is a two [present tense, as far as I can tell], what are [present tense] the odds that BOTH dice rolled [past tense] a two?"
People do not speak like this. People do not write like this. This would be edited in a technical writing assignment.
Let's match up the tenses, which is the way people speak, and the way non-quantum reality works. "If you rolled the dice, and at least one of the dice was a two, what were the odds that BOTH dice rolled a two?"
Mixing tenses (in a way that would get you edited in any journal) just to create confusion is childish. The lack of reference to the specific number of dice rolls has two effects. Math folks will interpret the lack of explication as "any number of rolls" or some abstract scenario. Non-math folks will tend to assume it refers to one roll of one pair of dice with no reference to probability in general.
In addition, the use of the present tense implies (in my mind) that one is talking about more than one roll. The use of past tense implies that it's more likely that one is talking about a single roll.
Last edited by redietz; 05-13-2015 at 04:53 PM.
Originally Posted by redietz
And what were the odds? If you think 1/6 cause the die has 6 faces, then why is that when you bet on that 1/6 event at a 8 for 1 odds you still lose?
At no point have I said the odds are/were (I can be as tricky as you boys) 1/6. Perhaps you should actually read my posts and test yourselves for reading comprehension as you would have others tested for math comprehension.
This trick question has been around a long time, decades I think, so don't believe you've discovered some great brain teaser.
Do tell what do you think the odds are then. Please.
I should probably be making fun of your reading comprehension. My take on this entire waste of space has been as crystal clear from the start as your trick question. Some people (arci, Rob) got it. Some people didn't.
Gee, I must have divided the reading audience into those who got it and dullards. Mercy me, what a rush (yes, Sheldon, that was sarcasm). What a clever boy I am.
Last edited by redietz; 05-13-2015 at 08:36 PM.
Hi PharoahsWin and thanks for joining and for posting.
I agree it's a trick question. But let's go back to the actual, original question:Originally Posted by PharoahsWin
You have two 6-sided dice in a cup. You shake the dice, and slam the cup down onto the table, hiding the result. Your partner peeks under the cup, and tells you, truthfully, "At least one of the dice is a 2." What is the probability that both dice are showing a 2?
The "trick" is really in how you decide to solve the problem. Clearly the two sides (1/11 vs 1/6) are divided in their methodology.
The 1/6 camp looks at the simple, in-your-face condition: one die shows a two, so what are the odds that the other die will also show a 2? That's 1/6.
The 1/11 camp doesn't look at the simple, in-your-face condition and looks at all of the different combinations of dice containing a 2 and then the single combination that has 2-2 which is 1/11.
I said it early on that it was a trick question and if you want to ignore the conditions of the question the answer is 1/36.
If you accept the condition of the question with one die already showing a 2 it's 1/6.
But if you don't accept the condition of the question and start questioning it by saying things like we don't know which of the two dice is a 2 then you will choose the 1/11 answer.
I accept that out of the two dice one is showing a 2. I don't care which of the two dice it is and with one die showing a 2 the only choice for me are the six sides on the other/second die.
Indeed how you read and interpret the question will define your answer. And because interpretations can be different there can be four answers:
1. 1/36
2. 1/11
3. 1/10
4. 1/6
What I said early on was that the question was written to throw a curve ball. I think the question was written to make you look for the 1/11 answer because of this "set-up line" that preceded the question:
Along the vein of the Two Coin Puzzle,
It was the two coin puzzle that asked you to look for the combinations of two coins, without one coin being "set" so you would get HH, TT, HT, TH.
But ignoring the line Along the vein of the Two Coin Puzzle, I accept that one die is set as 2 which removes the other five faces on that die and leaves only the six faces on the second die. When I start reading comments like we don't know which of the two dice is the 2 I just throw my hands up in disbelief. For heaven's sake -- there are only two dice and 2-1=1.
Clearly both the Wizard and "miplet" when they did their videos explaining the two dice problem they did exactly opposite what the original question stated as a condition and they rotated both dice to show the different combinations. Had they left one die frozen showing a 2 and did not touch that die they would be left with only one die to rotate for an answer of 1/6.
I've asked this over and over again: why rotate the dice known to be showing a 2? The original question told you "at least one of the dice is a 2" so why change that condition? And by rotating that die to explain the 1/11 answer they corrupted the question.
So, it was the 1/11 folks who were tricked. They were tricked into not following what the question asked. Whether or not that was the intent of the author of the question I don't know. But perhaps it was unintended because as redietz who is also a writer pointed out, the question was written very poorly. Very poorly. And as I said early in the discussion, had this been part of any national standardized test like an SAT or a LSAT it would have been thrown out just as another confusing SAT question was thrown out back in the 1960s.
Understanding that this is a trick question, look at what you posted.
The bold "other" in the quote there, is exactly what I was trying to explain to you in my first post. You're clearly saying that you feel that if one of the dice is a 2, then the "OTHER" die has a 1 in 6 chance of being a 2 as well. This is 100% correct, but the question is asking the odds of "BOTH" dice showing a 2, not the "OTHER" die.Originally Posted by Alan Mendelson
You understand, that there's 1 in 11 combinations that can produce a 2-2, when at least 1 die is a 2. This is, of course, before the dice are actually rolled. 11 combination with a 2, only 1 that is a 2-2, before you roll the dice. I know you understand this because you've said so on other posts. So, before the dice are rolled there's a 1 in 11 chance of 2-2. Then, the dice are rolled, the peeker peeks, and the question asks what are the odds "BOTH" dice rolled 2's. Tricky, but the answer is still 1 in 11.
As, the quote above shows, you change the question to "We know 1 die is a 2, so the "OTHER" die is 1 in 6 to be a 2. Which is absolutely correct, and understandable, if one die is a 2 the "OTHER" has a 1 in 6 chance of being a 2. But, this does not answer the question of the odds "BOTH" dice rolled 2's. There's a 1 in 11 chance that "BOTH" dice rolled 2's. 1 in 6 the "other" die rolled a 2.
I understand it's tricky, but if you look at all your responses, including the response to my post, you are answering what the odds of the "OTHER" die rolling a 2 is. This is your arguement, that the "OTHER" die is 1 in 6 to roll a 2. The question, however, clearly asks what the odds are "BOTH" dice rolled a 2.
This is why the wizards bet will lose you money. Why all the tests show 1 in 11. And, why if you try the experiment yourself you'll get 2-2 1 in 11 times, while the "other" die always has a 1 in 6 chance of being a 2.
The thing that really tripped you up, Alan, was taking that bet. The bet will lose you money because the answer is 1 in 11. Otherwise, 1 in 6, was a valid arguement, if you could get people to see the question the "other" way. But, taking that bet shows that you really didn't understand the question, because it would cost you and everyone else who takes the bet a lot of money.
You can test the bet out and see for yourself. 2-2 comes up 1 in 11 tries, even though the "other" die is 1 in 6 to throw a deuce. That's just never going to change, and it will always cost you money. Misunderstanding or not.
Last edited by PharoahsWin; 05-13-2015 at 10:51 PM.
As I have said over and over again: there are two dice and at least one of them is a 2. If you have two dice and at least one of them is a 2 whatever die you decide is a 2 has no other odds or options or choices. It is a 2 -- period.Originally Posted by PharoahsWin
Even if you think "at least one of the dice is a 2" means that both dice are showing a 2, you still would consider one of them and say "if dice A is a 2, dice B only has a 1/6 chance of being a 2 also."
Look math guys -- there are two ways to attack this problem: using rational, common sense, or creating the blue print for a Rube Goldberg machine.
And stop with your "test" of 1/11. Why don't you try MY TEST: Set one die as a 2 and then roll the other die and see if a 2 shows up 1/6 times?
The reason you have to do the 1 in 11 tests, is because that's the bet you made with the wizard. LOL. Honestly, I'm not trying to trick you, that's the bet you said you'd accept with the wizard. And, that bet will lose you a load of money, because the answer to that question is 1 in 11.
If you would have made a bet that the "other" die rolls a 2, 1 in 6 times, I would have left this whole thing alone. LOL. It's honestly what is tripping your whole arguement up. Unless you just like to lose money for some reason.
Alan, your test, asks the tester to roll just 1 die, not "BOTH" die. That 1 die, of course, being the "other" die. As I said, the "other" die is always 1 in 6 to be a deuce. Roll "Both" die, and you lose money.
Last edited by PharoahsWin; 05-13-2015 at 11:35 PM.
Just IMAGINE this scenario.
I have 60 dice and roll them all. I remove up to 10 of the dice that show a 2. If there are less than 10 deuces, I remove them all. If there are more than 10 deuces, I only remove 10. You do not know I did this, since you're in another room.
You walk in, can't see any of the dice. I tell you "pick one of the dice, any die you want" (they're all under Dixie cups, sorry forgot to mention). And now I ask you "Okay, what's the chance there's a deuce under that cup?"
From your perspective (without the info that I removed the "up to 10 deuces"), would say " 1/6 chance it's a deuce". Which is correct with the information you have.
We make our bet, I say I'll pay you 9-to-1 if it's a deuce, and you accept. I win your $1. Then I tell you, "oh yeah, I removed a bunch of the deuces". And you would (hopefully) say " well that's not fair!"
Does this illustration make sense to you? I'm not saying it's the same as the 2-dice problem. However, it is very similar.
Thoughts?
PharoahsWin: Yeah, well it was a good natured bet with a limited number of trials that I never expected to win and was limited to buying lunch because I wouldn't mind having lunch with Mike. Actually the bet was proposed by others in an attempt to monetize the debate. So, why not?
As I said on the WOV I'd be very happy to buy lunch for the Wizard and enjoy the time.
As we all know anything can happen rolling two dice but that's not the real issue. The real issue is how to read the question, and that is the great divide.
No I can't imagine that scenario. Try to imagine this one: You have two dice and at least one of them is a 2. Knowing that at least one of them is a two what does the other die have to show to have the combination 2-2? How many faces are on that other die? How many of those faces have the number 2?Originally Posted by RS__
Okay, then I'm gone. May I suggest asking the math guys... When at least one 2 is rolled, what are the odds the "other" die also rolled a 2?Originally Posted by Alan Mendelson
I bet they give your argument in reverse. I bet they'll say that the odds the "other" die rolls a 2 is 1 in 11, because if the "other" die rolled a 2, that would be a 2-2, which had a 1 in 11 chance of happening. So, they'd say, the "other" die had a 1 in 11 chance of being 2. They would be wrong, and you would be right saying 1 in 6.
Last edited by PharoahsWin; 05-14-2015 at 12:37 AM.
Pharoah when will you guys get a grip on reality: two dice have a total of 12 faces, right? When you have one die showing a 2 the other five faces on that die no longer count or matter or can be used. That leaves the six faces on the other die.
Your mathematical gymnastics of switching the 2 from one die to another sounds good for math theory, but when you have two physical, real dice, you can't do that.
So let me suggest this: There is no cup. This roll of two dice is done in the open.
You roll two dice. You observe "at least one of the two dice is a 2." Now tell me, can you really consider 1/11 as your answer?
It seems to me the only way you can possibly consider 1/11 as the answer because you don't actually see one die with a 2. But you don't really have to see one die has a two. You are told in the question that at least one die has a two.
And thanks for stopping by.
LOL. Alan, I wasn't trying to trick you there. LOL. Hasta.
Really Alan, you can't imagine that? What a poor imagination you must have...
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